- example
`display(lme)`

Akaike information criterion (AIC) is *AIC* =
–2*log*L*_{M} +
2*(*nc* + *p* + 1), where log*L*_{M} is
the maximized log likelihood (or maximized restricted log likelihood)
of the model, and *nc* + *p* + 1
is the number of parameters estimated in the model. *p* is
the number of fixed-effects coefficients, and *nc* is
the total number of parameters in the random-effects covariance excluding
the residual variance.

Bayesian information criterion (BIC) is *BIC* =
–2*log*L*_{M} +
ln(*n _{eff}*)*(

If the fitting method is maximum likelihood (ML), then

*n*=_{eff}*n*, where*n*is the number of observations.If the fitting method is restricted maximum likelihood (REML), then

*n*=_{eff}*n*–*p*.

A lower value of deviance indicates a better fit. As the value
of deviance decreases, both AIC and BIC tend to decrease. Both AIC
and BIC also include penalty terms based on the number of parameters
estimated, *p*. So, when the number of parameters
increase, the values of AIC and BIC tend to increase as well. When
comparing different models, the model with the lowest AIC or BIC value
is considered as the best fitting model.

`LinearMixedModel`

computes the deviance of
model *M* as minus two times the loglikelihood of
that model. Let *L*_{M} denote
the maximum value of the likelihood function for model *M*.
Then, the deviance of model *M* is

$$-2*\mathrm{log}{L}_{M}.$$

A lower value of deviance indicates a better fit. Suppose *M*_{1} and *M*_{2} are
two different models, where *M*_{1} is
nested in *M*_{2}. Then, the
fit of the models can be assessed by comparing the deviances *Dev*_{1} and *Dev*_{2} of
these models. The difference of the deviances is

$$Dev=De{v}_{1}-De{v}_{2}=2\left(\mathrm{log}L{M}_{2}-\mathrm{log}L{M}_{1}\right).$$

Usually, the asymptotic distribution of this difference has
a chi-square distribution with degrees of freedom *v* equal
to the number of parameters that are estimated in one model but fixed
(typically at 0) in the other. That is, it is equal to the difference
in the number of parameters estimated in M_{1} and
M_{2}. You can get the *p*-value
for this test using `1 – chi2cdf(Dev,V)`

,
where *Dev* = *Dev*_{2} – *Dev*_{1}.

However, in mixed-effects models, when some variance components fall on the boundary of the parameter space, the asymptotic distribution of this difference is more complicated. For example, consider the hypotheses

*H*_{0}: $$D=\left(\begin{array}{cc}{D}_{11}& 0\\ 0& 0\end{array}\right),$$ *D* is a *q*-by-*q* symmetric
positive semidefinite matrix.

*H*_{1}: *D* is
a (*q*+1)-by-(*q*+1) symmetric positive
semidefinite matrix.

That is, *H*_{1} states
that the last row and column of *D* are different
from zero. Here, the bigger model *M*_{2} has *q* +
1 parameters and the smaller model *M*_{1} has *q* parameters.
And *Dev* has a 50:50 mixture of *χ*^{2}_{q} and *χ*^{2}_{(q +
1)} distributions (Stram and Lee, 1994).

[1] Hox, J. *Multilevel Analysis, Techniques and
Applications*. Lawrence Erlbaum Associates, Inc., 2002.

[2] Stram D. O. and J. W. Lee. "Variance components
testing in the longitudinal mixed-effects model". *Biometrics*,
Vol. 50, 4, 1994, pp. 1171–1177.

Was this topic helpful?